Bài 1:

\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)

\(\left(x+1\right)\left(x+2\right)-\left(x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+1-x-3\right)=0\)

\(\Leftrightarrow-2\left(x+2\right)=0\)

\(\Leftrightarrow x=-2\)

\(4x^2-12x+5=4x^2-10x-2x+5=2x\left(2x-5\right)-\left(2x-5\right)=\left(2x-1\right)\left(2x-5\right)\)

\(\left(x+1\right)\left(x+2\right)-\left(x+2\right)\left(x+3\right)=0\)

\(\left(x+2\right)\left(-2\right)=0\)\(\Rightarrow x+2=0\) hay \(x=-2\)

\(4x^2-12x+5=4\left(x^2-3x+\frac{5}{4}\right)=4\left(x^2-3x+\frac{9}{4}-\frac{9}{4}+\frac{5}{4}\right)=4\left(x-\frac{3}{2}\right)^2-2=\left(2x-1\right)\left(2x-5\right)\)

\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)

1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)

2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)

\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)=\left(3x-2y\right)\left(3x+2y+2\right)\)

3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)

1) Ta có: \(x^3+2x^2-6x-27\)

\(=\left(x-3\right)\left(x^2+3x+9\right)+2x\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

2: Ta có: \(9x^2+6x-4y^2-4y\)

\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)\)

\(=\left(3x-2y\right)\left(3x+2y+2\right)\)

a)  \(4x^2+20x+25=\left(2x+5\right)^2\)

b) \(x^2-6x+9=\left(x-3\right)^2\)

c) \(4x^2+12x+9=\left(2x+3\right)^2\)

a)  \(27x^3-8=\left(3x-2\right)\left(9x^2+6x+4\right)\)

b)  \(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)

c)  \(\left(2y-1\right)^1-4x^2+4x-1=\left(2y-1\right)^2-\left(2x-1\right)^2=\left(2y-1-2x+1\right)\left(2y-1+2x-1\right)\)

\(=\left(2y-2x\right)\left(2y+2x-2\right)=4\left(y-x\right)\left(y+x-1\right)\)

a)\(2x^2-12x=-18\)

\(\Leftrightarrow2x^2-12x+18=0\)

\(\Leftrightarrow2\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow2\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)^2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

b) \(\left(4x^2-4x+1\right)-x^2=0\)

\(\Leftrightarrow\left(2x-1\right)^2-x^2=0\)

\(\Leftrightarrow\left(2x-1-x\right)\left(2x-1+x\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}}\)

_Minh ngụy_

\(x^2-ay-y^2-ax\)

\(=\left(x^2-y^2\right)-\left(ax+ay\right)\)

\(=\left(x-y\right)\left(x+y\right)-a\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-a\right)\)

_Minh ngụy_

a) \(8x\left(x-3\right)+x-3=0\)

\(\Rightarrow8x\left(x-3\right)+\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(8x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{8}\end{matrix}\right.\)

b) \(x^2+36=12x\)

\(\Rightarrow x^2-12x+36=0\)

\(\Rightarrow\left(x-6\right)^2=0\)

\(\Rightarrow x=6\)

a)\(4x^2-4x+4=0\Leftrightarrow\left(2x-1\right)^2+3\) (đến đây hết pt dc rùi)

b)\(x^3-27=\left(x-3\right)\left(x^2+3x+9\right)\)

c)\(x^3-4x^2+3x=x^3-x^2-3x^2+3x\)

                                   =\(x^2\left(x-1\right)-3x\left(x-1\right)\)

                                 =\(x\left(x-3\right)\left(x-1\right)\)

d)\(4x^2-12x+3=\left(2x-3\right)^2-6\)

                                    =\(\left(2x-3\right)^2-\sqrt{6^2}\)

                                 =\(\left(2x-3-\sqrt{6}\right)\left(2x-3+\sqrt{6}\right)\)

\(a,4x^2-4x+4=4\left(x^2-x+1\right)\)

\(b,x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+9\right)\)

\(c,x^3-4x^2+3x=x\left(x^2-4x+3\right)\)

                             \(=x\left[\left(x^2-x\right)-\left(3x-3\right)\right]\)

                             \(=x\left[x\left(x-1\right)-3\left(x-1\right)\right]\)

                             \(=x\left(x-1\right)\left(x-3\right)\)

\(d,4x^2-12x+3=4\left(x^2-3x+\frac{3}{4}\right)\)

\(=4\left(x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{9}{4}+\frac{3}{4}\right)\)

\(=4\left[\left(x-\frac{3}{2}\right)^2-\frac{3}{2}\right]\)

\(=4\left[\left(x-\frac{3}{2}\right)^2-\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^2\right]\)

\(=4\left(x-\frac{3}{2}-\frac{\sqrt{3}}{\sqrt{2}}\right)\left(x-\frac{3}{2}+\frac{\sqrt{3}}{\sqrt{2}}\right)\)

\(=4\left(x-\frac{3+\sqrt{6}}{2}\right)\left(x-\frac{3-\sqrt{6}}{2}\right)\)

P/s: Dương: câu d t k chắc nx, sai thì thông cảm :)) -Huyền Nhi-